Problem: Solve for $x$ and $y$ by deriving an expression for $y$ from the second equation, and substituting it back into the first equation. $\begin{align*}-8x-2y &= 6 \\ -3x-2y &= -9\end{align*}$
Solution: Begin by moving the $x$ -term in the second equation to the right side of the equation. $-2y = 3x-9$ Divide both sides by $-2$ to isolate $y$ $y = {-\dfrac{3}{2}x + \dfrac{9}{2}}$ Substitute this expression for $y$ in the first equation. $-8x-2({-\dfrac{3}{2}x + \dfrac{9}{2}}) = 6$ $-8x + 3x - 9 = 6$ Simplify by combining terms, then solve for $x$ $-5x - 9 = 6$ $-5x = 15$ $x = -3$ Substitute $-3$ for $x$ back into the top equation. $-8( -3)-2y = 6$ $24-2y = 6$ $-2y = -18$ $y = 9$ The solution is $\enspace x = -3, \enspace y = 9$.